Hi! if u=2xy, then du/dx=?
Wolfram says use product rule but I thought the answer was just 2y...thanks!:O)

Question

Hi! if u=2xy, then du/dx=?
Wolfram says use product rule but I thought the answer was just 2y...thanks!:O)

anonymous · Answer

Y is a completely new variable so you have to treat it as such. Since you're taking the derivative with respect to x, x, can cancel out, but y is replaced with y prime (y')

anonymous · Answer

So the product rule is as follows:
(dy/dx) F * G = F' * G + F * G'

anonymous · Answer

@undeadknight26

anonymous · Answer

Thusly,
(du/dx) 2xy = 2y + 2xy'

anonymous · Answer

Does that make sense? You're using implicit differentiation

anonymous · Answer

yikes...I can't believe I forgot this!
so if u=2xy+xy^2 then du/dx would be y+x(dy/dx)+2xy(dy/dx) ?

anonymous · Answer

do I have this correct?

anonymous · Answer

You'd need to use the product rule for xy^2 as well

anonymous · Answer

didn't I?

anonymous · Answer

I thought I did

anonymous · Answer

It seems that you only used the product rule for 2xy

anonymous · Answer

and even so, you left off the 2 constant

anonymous · Answer

opps...one sec...

anonymous · Answer

if u=2xy+xy^2 then du/dx=[y+x(dy/dx)] + [y^2 + 2xy(dy/dx)]  ?
How's that?

anonymous · Answer

Don't forget your y prime!!
(dy/dx) yx = y + xy'

anonymous · Answer

so did I do that correct this time?

anonymous · Answer

if u=2xy+xy^2 then du/dx=[y+x(dy/dx)] + [y^2 + 2xy(dy/dx)] ?

anonymous · Answer

Once you take the derivative, you no longer have to use the term (dy/dx)

anonymous · Answer

oh, I see. You're using the term (dy/dx) instead of just writing y'

anonymous · Answer

dy/dx is the same as typing y' right?

anonymous · Answer

Yeah, but I'm used to seeing y' for y prime xD

anonymous · Answer

It looks good to me!! :)

anonymous · Answer

lemme try once more lol
if u=2xy+xy^2 then du/dx=[y+xy'] + [y^2 + 2xyy'] ?

anonymous · Answer

that is much faster to type for sure!...thanks for the help! I can't believe I occasionally forget these small dumb little things...geez! :O)

anonymous · Answer

WAIT, you forgot the 2 constant for [y+xy']

anonymous · Answer

opps...uhg...ok, one last time...lol

anonymous · Answer

if u=2xy+xy^2 then du/dx=[2y+2x(dy/dx)] + [y^2 + 2xy(dy/dx)] ?

anonymous · Answer

good?

anonymous · Answer

PERFECT! :D

anonymous · Answer

whew...thanks Aeg...back to studying now...here's your cookie! Thanks again!